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2 years ago

For the common-emitter, common-base and emitter-follower amplifier designs,what is the primary benefit of each amplifier?

Physics

1 answer:

kati45 [8]2 years ago

6 0

Answer:

Explanation:

A common emitter amplifier works by inverting. It does have low input impedance. Despite its low input impedance, it poses an otherwise high output impedance.

The common base circuit performs optimally when it acts as a current buffer. It has the ability to take an input current at a low input impedance, and transmit almost the same current to an impedance with a higher output

The primary benefit of emitter follower amplifier is that the transistor is able to provide current and power gain. Although this transistor takes in little current from the input. It still provides an impedance with a low output to a circuit by exercising the output of the follower. This then translates to that the output under load not dropping.

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__________ theory argues that city growth is generated by the pressure from the city center to expand outward. Expansion threate

artcher [175]

Answer: Concentric zone

Explanation:

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2 years ago

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A car is initially driving at 30 m/s. It hits a large pothole, after which it is traveling in the same direction but at 25 m/s.

Bogdan [553]

Answer:

The time of Mars is 1.65 times larger on Mars than on Earth

Explanation:

The equation that describes the system is the final speed is equal to the speed minos the speed lost by the collision with the porhole

Vf = Vo - V pothole

B) let's transform the weight of free groin system and N international system

1 N = 0.2248 lb

2.8 lbs (1N / 0.2248lbs) = 12.5 N

c) Kinematic equations are the same in all inertial systems, Mars and Earth, so we can use the height equation, with zero initial velocity

Y = Vo t - ½ g t²

Y = - ½ g t²

t = √ 2Y / g

Mars

gm = 0.37g

gm = 0.37 9.8

gm = 3,626 m / s²

t = √( 2 1.9 / 3.626 )

t = 1.02 s

Earth

t = √( 2 1.9 / 9.8)

t = 0.62 s

To make the comparison of time we are the relationship between the two

tm / te = 1.02 / 0.62

tm / te = 1.65

The time of Mars is 1.65 times larger on Mars than on Earth

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2 years ago

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A very strong, but inept, shot putter puts the shot straight up vertically with an initial velocity of 11.0 m/s. how long does h

noname [10]

The shot putter should get out of the way before the ball returns to the launch position.

Assume that the launch height is the reference height of zero.
u = 11.0 m/s, upward launch velocity.
g = 9.8 m/s², acceleration due to gravity.

The time when the ball is at the reference position (of zero) is given by
ut - (1/2)gt² = 0
11t - 0.5*9.8t² = 0
t(11 - 4.9t) = 0
t = 0 or t = 4.9/11 = 0.45 s

t = 0 corresponds to when the ball is launched.
t = 0.45 corresponds to when the ball returns to the launch position.

Answer: 0.45 s

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2 years ago

Two clouds collide and form another, more massive cloud. One cloud is stationary, while the other is traveling at 1 m/s. After t

sweet [91]

Answer: 3

Explanation:

Given

One cloud is traveling at rate of $u_2=1\ m/s$

combined velocity of the two is $v=0.25\ m/s$

Suppose the masses of the clouds be $m_1,m_2$

Conserving momentum

$\Rightarrow m_1u_1+m_2u_2=\left(m_1+m_2\right)v\\\text{Divide whole equation by }m_2\\\Rightarrow \dfrac{m_1}{m_2}u_1+u_2=\left(\dfrac{m_1}{m_2}+1\right)v\\\\\Rightarrow 0+1=0.25\dfrac{m_1}{m_2}+0.25\\\\\Rightarrow \dfrac{m_1}{m_2}=\dfrac{0.75}{0.25}\\\\\Rightarrow \dfrac{m_1}{m_2}=3$

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2 years ago

xConsider the following reduction potentials: Cu2+ + 2e– Cu E° = 0.339 V Pb2+ + 2e– Pb E° = –0.130 V For a galvanic cell employi

slega [8]

Answer:

Approximately $\rm 90\; kJ$ .

Explanation:

Cathode is where reduction takes place and anode is where oxidation takes place. The potential of a electrochemical reaction ( $E^{\circ}(\text{cell})$ ) is equal to

$E^{\circ}(\text{cell}) = E^{\circ}(\text{cathode}) - E^{\circ}(\text{anode})$ .

There are two half-reactions in this question. $\rm Cu^{2+} + 2\,e^{-} \rightleftharpoons Cu$ and $\rm Pb^{2+} + 2\,e^{-} \rightleftharpoons Pb$ . Either could be the cathode (while the other acts as the anode.) However, for the reaction to be spontaneous, the value of $E^{\circ}(\text{cell})$ should be positive.

In this case, $E^{\circ}(\text{cell})$ is positive only if $\rm Cu^{2+} + 2\,e^{-} \rightleftharpoons Cu$ is the reaction takes place at the cathode. The net reaction would be

$\rm Cu^{2+} + Pb \to Cu + Pb^{2+}$ .

Its cell potential would be equal to $0.339 - (-0.130) = \rm 0.469\; V$ .

The maximum amount of electrical energy possible (under standard conditions) is equal to the free energy of this reaction:

$\Delta G^{\circ} = n \cdot F \cdot E^{\circ} (\text{cell})$ ,

where

$n$ is the number moles of electrons transferred for each mole of the reaction. In this case the value of $n$ is $2$ as in the half-reactions.
$F$ is Faraday's Constant (approximately $96485.33212\; \rm C \cdot mol^{-1}$ .)

$\begin{aligned}\Delta G^{\circ} &= n \cdot F \cdot E^{\circ} (\text{cell})\cr &= 2\times 96485.33212 \times (0.339 - (-0.130)) \cr &\approx 9.0 \times 10^{4} \; \rm J \cr &= 90\; \rm kJ\end{aligned}$ .

5 0

2 years ago